I have trouble understanding the proof of Proposition 2.18 in Karatzas/Shreve: Brownian Motion and Stochastic Calculus, which states that a $\mathcal{F_t}$-progressively measurable process $X_t$, stopped at a $\mathcal{F_t}$-stopping time $\tau$, remains progressive, i. e. $X_{t\wedge\tau}$ is $\mathcal{F_t}$-progressive.
The main idea is to decompose the map $[0,T]\times\Omega\to\mathbb{R}^d, (t,\omega)\mapsto X_{t\wedge\tau(\omega)}(\omega)$ into $M: (t,\omega)\mapsto (t\wedge\tau(\omega),\omega)$ and $(t,\omega)\mapsto X_t(\omega)$ (for all $T>0$). Then it suffices to show that $M$ is $(\mathcal{B}[0,T]\otimes \mathcal{F}_T)$-measurable, since the other map is measurable because of the assumption of progressive measurability of $X$.
The main idea is clear to me. But I don’t see how $M$ is measurable. It seems intuitively clear to me (thinking of the filtration as information), but I don’t manage to prove it formally.